ac.commutative algebra - Locally square implies square

OK, I've got it. There is no such local criterion for squareness.

Let $k$ be a field of characteristic not $2$. Take the ring of triples $(f,g,h) in k[t]^3$, subject to the conditions that $f(1)=g(-1)$, $g(1)=h(-1)$ and $h(1)=f(-1)$. Consider the element $(t^2,t^2,t^2)$. If this were a square, its square root would have to be $(pm t, pm t, pm t)$. But two of those $pm$'s would be the same sign, and $t$ evaluated at $1$ and at $-1$ are not equal.

Now, to check that $(t^2, t^2, t^2)$ is everywhere locally a square. Geometrically, we are talking about three lines glued into a triangle. Any prime ideal has a neighborhood which is contained in the union of two neighboring lines, say the first two. On the first two lines, $(-t, t, 1)$ is a square root of $(t^2, t^2, t^2)$.

For the suspicious, an algebraic proof. Set $u_1=(0, (1+t)/2, (1-t)/2)$ and let $u_2$ and $u_3$ be the cyclic permutations thereof. We have $u_1+u_2+u_3=1$ so, in any local ring, one of the $u_i$ must be a unit. WLOG, suppose that $u_1$ is a unit. Notice that $u_1 (1, -t, t)^2 = u_1 (t^2, t^2, t^2)$. So, in a local ring where $u_1$ is a unit, $(1,t,-t)$ is a square root of $(t^2, t^2, t^2)$.